我有桌子:
id | name
1 | a,b,c
2 | b
我想要这样的输出:
id | name
1 | a
1 | b
1 | c
2 | b
答案 0 :(得分:87)
如果您可以创建一个数字表,其中包含从1到要拆分的最大字段的数字,您可以使用这样的解决方案:
select
tablename.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(tablename.name, ',', numbers.n), ',', -1) name
from
numbers inner join tablename
on CHAR_LENGTH(tablename.name)
-CHAR_LENGTH(REPLACE(tablename.name, ',', ''))>=numbers.n-1
order by
id, n
请参阅小提琴here。
如果您无法创建表格,那么解决方案可以是:
select
tablename.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(tablename.name, ',', numbers.n), ',', -1) name
from
(select 1 n union all
select 2 union all select 3 union all
select 4 union all select 5) numbers INNER JOIN tablename
on CHAR_LENGTH(tablename.name)
-CHAR_LENGTH(REPLACE(tablename.name, ',', ''))>=numbers.n-1
order by
id, n
示例小提琴是here。
答案 1 :(得分:5)
<强> I have take the reference from here with changed column name. 强>
DELIMITER $$
CREATE FUNCTION strSplit(x VARCHAR(65000), delim VARCHAR(12), pos INTEGER)
RETURNS VARCHAR(65000)
BEGIN
DECLARE output VARCHAR(65000);
SET output = REPLACE(SUBSTRING(SUBSTRING_INDEX(x, delim, pos)
, LENGTH(SUBSTRING_INDEX(x, delim, pos - 1)) + 1)
, delim
, '');
IF output = '' THEN SET output = null; END IF;
RETURN output;
END $$
CREATE PROCEDURE BadTableToGoodTable()
BEGIN
DECLARE i INTEGER;
SET i = 1;
REPEAT
INSERT INTO GoodTable (id, name)
SELECT id, strSplit(name, ',', i) FROM BadTable
WHERE strSplit(name, ',', i) IS NOT NULL;
SET i = i + 1;
UNTIL ROW_COUNT() = 0
END REPEAT;
END $$
DELIMITER ;
答案 2 :(得分:3)
我的变体:将表名,字段名和分隔符作为参数的存储过程。灵感来自帖子http://www.marcogoncalves.com/2011/03/mysql-split-column-string-into-rows/
delimiter $$
DROP PROCEDURE IF EXISTS split_value_into_multiple_rows $$
CREATE PROCEDURE split_value_into_multiple_rows(tablename VARCHAR(20),
id_column VARCHAR(20), value_column VARCHAR(20), delim CHAR(1))
BEGIN
DECLARE id INT DEFAULT 0;
DECLARE value VARCHAR(255);
DECLARE occurrences INT DEFAULT 0;
DECLARE i INT DEFAULT 0;
DECLARE splitted_value VARCHAR(255);
DECLARE done INT DEFAULT 0;
DECLARE cur CURSOR FOR SELECT tmp_table1.id, tmp_table1.value FROM
tmp_table1 WHERE tmp_table1.value IS NOT NULL AND tmp_table1.value != '';
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;
SET @expr = CONCAT('CREATE TEMPORARY TABLE tmp_table1 (id INT NOT NULL, value VARCHAR(255)) ENGINE=Memory SELECT ',
id_column,' id, ', value_column,' value FROM ',tablename);
PREPARE stmt FROM @expr;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
DROP TEMPORARY TABLE IF EXISTS tmp_table2;
CREATE TEMPORARY TABLE tmp_table2 (id INT NOT NULL, value VARCHAR(255) NOT NULL) ENGINE=Memory;
OPEN cur;
read_loop: LOOP
FETCH cur INTO id, value;
IF done THEN
LEAVE read_loop;
END IF;
SET occurrences = (SELECT CHAR_LENGTH(value) -
CHAR_LENGTH(REPLACE(value, delim, '')) + 1);
SET i=1;
WHILE i <= occurrences DO
SET splitted_value = (SELECT TRIM(SUBSTRING_INDEX(
SUBSTRING_INDEX(value, delim, i), delim, -1)));
INSERT INTO tmp_table2 VALUES (id, splitted_value);
SET i = i + 1;
END WHILE;
END LOOP;
SELECT * FROM tmp_table2;
CLOSE cur;
DROP TEMPORARY TABLE tmp_table1;
END; $$
delimiter ;
用法示例(规范化):
CALL split_value_into_multiple_rows('my_contacts', 'contact_id', 'interests', ',');
CREATE TABLE interests (
interest_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
interest VARCHAR(30) NOT NULL
) SELECT DISTINCT value interest FROM tmp_table2;
CREATE TABLE contact_interest (
contact_id INT NOT NULL,
interest_id INT NOT NULL,
CONSTRAINT fk_contact_interest_my_contacts_contact_id FOREIGN KEY (contact_id) REFERENCES my_contacts (contact_id),
CONSTRAINT fk_contact_interest_interests_interest_id FOREIGN KEY (interest_id) REFERENCES interests (interest_id)
) SELECT my_contacts.contact_id, interests.interest_id
FROM my_contacts, tmp_table2, interests
WHERE my_contacts.contact_id = tmp_table2.id AND interests.interest = tmp_table2.value;
答案 3 :(得分:1)
CREATE PROCEDURE `getVal`()
BEGIN
declare r_len integer;
declare r_id integer;
declare r_val varchar(20);
declare i integer;
DECLARE found_row int(10);
DECLARE row CURSOR FOR select length(replace(val,"|","")),id,val from split;
create table x(id int,name varchar(20));
open row;
select FOUND_ROWS() into found_row ;
read_loop: LOOP
IF found_row = 0 THEN
LEAVE read_loop;
END IF;
set i = 1;
FETCH row INTO r_len,r_id,r_val;
label1: LOOP
IF i <= r_len THEN
insert into x values( r_id,SUBSTRING(replace(r_val,"|",""),i,1));
SET i = i + 1;
ITERATE label1;
END IF;
LEAVE label1;
END LOOP label1;
set found_row = found_row - 1;
END LOOP;
close row;
select * from x;
drop table x;
END
答案 4 :(得分:1)
这是我的尝试: 第一个选择将csv字段呈现给拆分。 使用递归CTE,我们可以创建一个数字列表,该列表限于csv字段中的术语数。 项的数量只是csv字段长度和其本身(所有定界符已删除)之间的差。 然后将这些数字与substring_index一起提取该术语。
with recursive
T as ( select 'a,b,c,d,e,f' as items),
N as ( select 1 as n union select n + 1 from N, T
where n <= length(items) - length(replace(items, ',', '')))
select distinct substring_index(substring_index(items, ',', n), ',', -1)
group_name from N, T
答案 5 :(得分:0)
最初的问题是针对MySQL和SQL的。下面的示例适用于新版本的MySQL。不幸的是,无法在任何SQL Server上运行的通用查询。有些服务器不支持CTE,另一些服务器则没有substring_index,而另一些服务器则具有内置功能,可将字符串分成多行。
---答案如下---
当服务器不提供内置功能时,递归查询很方便。它们也可能是瓶颈。
以下查询是在MySQL 8.0.16版上编写和测试的。它不适用于5.7-版本。旧版本不支持通用表表达式(CTE),因此不支持递归查询。
with recursive
input as (
select 1 as id, 'a,b,c' as names
union
select 2, 'b'
),
recurs as (
select id, 1 as pos, names as remain, substring_index( names, ',', 1 ) as name
from input
union all
select id, pos + 1, substring( remain, char_length( name ) + 2 ),
substring_index( substring( remain, char_length( name ) + 2 ), ',', 1 )
from recurs
where char_length( remain ) > char_length( name )
)
select id, name
from recurs
order by id, pos;
答案 6 :(得分:0)
如果name
列是JSON数组(例如'["a","b","c"]'
),则可以使用JSON_TABLE()提取/解压缩它:
select t.id, j.name
from mytable t
join json_table(
t.name,
'$[*]' columns (name varchar(50) path '$')
) j;
结果:
| id | name |
| --- | ---- |
| 1 | a |
| 1 | b |
| 1 | c |
| 2 | b |
如果您以简单的CSV格式存储值,则首先需要将其转换为JSON:
select t.id, j.name
from mytable t
join json_table(
concat('[', replace(json_quote(t.name), ',', '","'), ']'),
'$[*]' columns (name varchar(50) path '$')
) j;
结果
| id | name |
| --- | ---- |
| 1 | a |
| 1 | b |
| 1 | c |
| 2 | b |
答案 7 :(得分:0)
最佳做法。 结果:
SELECT
SUBSTRING_INDEX(SUBSTRING_INDEX('ab,bc,cd',',',help_id+1),',',-1) AS oid
FROM
(
SELECT @xi:=@xi+1 as help_id from
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc1,
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc2,
(SELECT @xi:=-1) xc0
) a
WHERE
help_id < LENGTH('ab,bc,cd')-LENGTH(REPLACE('ab,bc,cd',',',''))+1
首先,创建一个数字表:
SELECT @xi:=@xi+1 as help_id from
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc1,
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc2,
(SELECT @xi:=-1) xc0;
| help_id |
| --- |
| 0 |
| 1 |
| 2 |
| 3 |
| ... |
| 24 |
第二,只需拆分str:
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX('ab,bc,cd',',',help_id+1),',',-1) AS oid
FROM
numbers_table
WHERE
help_id < LENGTH('ab,bc,cd')-LENGTH(REPLACE('ab,bc,cd',',',''))+1
| oid |
| --- |
| ab |
| bc |
| cd |
答案 8 :(得分:-1)
这是我的解决方法
SELECT COUNT(*) FROM ${table.name}